A capacitor has parallel plates of area 10 cm 2 separated by 1.8 mm. The space b

Question

A capacitor has parallel plates of area 10 cm2 separated by 1.8 mm. The space between the plates is filled with polycarbonate (see the table below).

(a) Find the permittivity of polycarbonate.

(b) Find the maximum permissible voltage across the capacitor to avoid dielectric breakdown.

(c) When the voltage equals the value found in part (b), find the surface-charge density on each plate and the induced surface-charge density on the surface of the dielectric.

Explanation / Answer

let e = epsillon

1)We know e = ke0

e = 2.8*8.85*10^-12 = 24.78*10^-12

2) we also know that V = Em * d = 3*10^7 * 0.0018 = 5.4* 10^ -3 m/s

3) E= sigma / epsilon = V/d

sigma = e V / d = 2.4* 10^ -11 * 5.4*10^-3 / 0.0018 = 7.2*-10^-11

and

b) we also know sigma = e0 V /d

sigma = 8.85*10^ 12 *5.4*10^-3 / 0.0018 = 2.6*10^ -11

now, 7.2*10^ -11 - 2.6*10^-11 = 4.6*10^-11

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